Hello, for this problem , i am more concerned with the letter E. I know how it s

Question

Hello, for this problem , i am more concerned with the letter E. I know how it shifts but i do not know how to answer for E, i will give full points for anyone that can explain precisly and right...thank you!!

Consider the exothermic reaction: Cl20(g) + H2O(l) 2HCI(g) + O2(g). At equilibrium, 0.10 mol Cl20, 5.00 mol O2/ 10.0 mol HCI, and 2.00 mol H2O are present in a 20.0 L container at 25DegreeC. Do you need to consider the volume occupied by liquid H2O? Justify your answer. Calculate the value for Kc at 25DegreeC for the reaction shown above. Calculate the value for Kp at 25DegreeC for the reaction shown above. If the system described by the equation above is initially at equilibrium, predict the direction of the shift in the position of equilibrium (left, right, no effect, or cannot be determined) and justify your prediction (be complete) for each of the following. Temperature and container volume remain constant, except where noted. Some gaseous HCI is added. Some gaseous Cl2O removed. From the container is removed 0.50 mol liquid water. Container volume is halved; temperature remains constant. Temperature is decreased by 10*C; container volume remains constant. To the container is added 1.0 mol gaseous argon. To the container is added 1.0 g of solid rhodium to act as a catalyst. Container volume is doubled and temperature is raised by 10DegreeC. Some gaseous 02 is added and container volume is halved; temperature remains constant. For each of the changes shown in Part d above, what effect (decrease, increase, no effect, or cannot be determined) would there be on the value for the equilibrium constant for the reaction?

Explanation / Answer

a) No, since vol occupied by liquid is negligible as that compared to a gas. Rest all are gases, Hence vol of liquid can be neglected

b) Kc

[O2] = 5 / 20 = 0.25 M

[HCl] = 10/20 = 0.5

[Cl2O] = 0.1 / 20 = 0.005 M

Kc = 0.25 x 0.5^2 / 0.005 = 12.5

c) Kp = Kc (RT)^2

=> Kp = 12.5 x (0.0821 x 298)^2 = 7482,19

d) i) After addition of HCl equilibrium would shift left (according to le chateliers principle)

ii) removal of Cl2O would also shift the equilibrium to left

iii) removal of liquid water would shift the eq. to left

iv) If volume if halved, total P is doubled. Reaction moves in the direction of decreasing pressure. Hence reaction move back, LEFT.

v) T decreases, Pressure decreases => reaction moves in direction of increasig pressure, therefore moves right

vi) addition of 1 mol of argon. No effect

vii) No effect on eq. Catalyst helps attain equilibrium fast but does not affect the composition of equilibrium

viii) Final result of doubling vol. and increasing T is decrease in Pressure. Hence reaction shifts in forward direction

ix) moves left

e) No effest on eq. constant

Related Questions

Navigate

• Browse (All)
• Browse H
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.